Mathematics - Sequence & Series Question with Solution | TestHub

The given series is $S={\sum }_{n=1}^{\infty }\frac{n^2+n+1}{n!}$. We rewrite the numerator $n^2+n+1$ using the identity $n^2+n+1=n(n-1)+2n+1$. So, the $n$ -th term is $T_n=\frac{n(n-1)+2n+1}{n!}=\frac{n(n-1)}{n!}+\frac{2n}{n!}+\frac{1}{n!}$. For $n\ge 2$, $\frac{n(n-1)}{n!}=\frac{1}{(n-2)!}$. For $n\ge 1$, $\frac{2n}{n!}=\frac{2}{(n-1)!}$. The term $\frac{1}{n!}$ remains as is. Thus, $T_n=\frac{1}{(n-2)!}+\frac{2}{(n-1)!}+\frac{1}{n!}$ for $n\ge 2$. For $n=1$, $T_1=\frac{1^2+1+1}{1!}=3$. We can write $S=T_1+{\sum }_{n=2}^{\infty }{T}_n$.$$S=3+\sum _{n=2}^{\infty }\frac{1}{(n-2)!}+\sum _{n=2}^{\infty }\frac{2}{(n-1)!}+\sum _{n=2}^{\infty }\frac{1}{n!}$$ We know the series for $e={\sum }_{k=0}^{\infty }\frac{1}{k!}=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dots$. Using this: 1. ${\sum }_{n=2}^{\infty }\frac{1}{(n-2)!}=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dots =e$. 2. ${\sum }_{n=2}^{\infty }\frac{2}{(n-1)!}=2\left(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots \right)=2\left(e-\frac{1}{0!}\right)=2(e-1)$. 3. ${\sum }_{n=2}^{\infty }\frac{1}{n!}=\left(\frac{1}{2!}+\frac{1}{3!}+\dots \right)=e-\left(\frac{1}{0!}+\frac{1}{1!}\right)=e-(1+1)=e-2$. Adding these components with $T_1$: $S=3+e+2(e-1)+(e-2)=3+e+2e-2+e-2=4e-1$.